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18.034 Honors Differential Equations Spring 2009

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LECTURE 4. SEPARABLE EQUATIONS

Separable equations. Separable equations are differential equations of the form

dy f(x)(4.1) = .

dx g(y)

For example, x + yy� = 0 and y� = y2 − 1. A separable equation (4.1) can be written in the differential form as

(4.2) f(x)dx = g(y)dy.

Then, it can be solved formally by integrating both sides of (4.2). We state and prove the rigorous theory of local solutions for (4.2) (and hence (4.1)).

Theorem 4.1. Let f(x) and g(x) be continuous in the rectangle R = {(x, y) : a < x < b, c < y < d}. In addition, if f and g do not vanish simultaneously at any point of R, then (4.2) has one and only one solution through each point (x0, y0) ∈ R. The solution is given by

x y

(4.3) f(x)dx = g(y)dy x0 y0

It is essential that f and g do not vanish simultaneously. For example, xdx = −ydy has no solution through the origin.

Proof. Note that f(x) =� 0 for a < x < b or g(y) = 0 � for c < y < d. Without loss of generality, we assume g(y) > 0 for c < y < d.

Let � � x y

F (x) = f(x)dx, G(x) = g(y)dy x0 y0

so that (4.3) becomes

(4.4) F (x) = G(x).

Since G�(y) = g(y) > 0 in R, by the inverse function theorem, G−1 exists and (4.4) can be written as y = G−1(F (x)). That means, dy exists. Then, by differentiating (4.4), we get dx

F �(x) = G�(x) dy

, or f(x) = g(y) dy

. dx dx

This implies (4.2). Moreover, (4.3) gives the initial condition that y = y0 when x = x0. To prove the uniqueness, let y be one solution of (4.2) and z be another solution with the same

initial condition. Under the hypothesis, the equation

dz =

f(x) dx g(z)

implies that dz/dx exists for any (x, z) ∈ R. Let

u = G(y), v = G(z).

Then, du dy dy

= G�(y) = g(y) = f(x). dx dx dx

1

� �

dzSimilarly, = f(x). Since u and v have the same derivative, they differ by a constant. On the

dx other hand, the initial conditions for u and v at x0 agree. Therefore, u = v everywhere in R. This completes the proof. �

Example 4.2. Consider the initial value problem

(4.5) dy

= 1 + y 2 , y(0) = 1. dx

Separating the variables, we write the differential equation as dy

= dx.1 + y2

Since the constant function never vanishes, upon integration and evaluation, we obtain

tan−1 y = x + c, tan−1 1 = c.

Therefore, the (unique) solution of (4.5) is y = tan(x + π/4). The same result is obtained by integrating between corresponding limits

y dy x

= dx.1 + y2

1 0

Orthogonal trajectories. If two families of curves are such that every curve of one family inter­sects the curves of the other family at a right angle, then we say that the two families are orthogonal trajectories of each other. For example, the coordinate lines:

x = c1, y = c2

in a Cartesian coordinate system form a set of orthogonal trajectories. Another example is the circles and radial lines

r = c1, θ = c2

in a polar coordinate system. Suppose a curve in the (x, y)-plane is such that the tangent at a point (x, y) on it makes an angle

φ with the x-axis. The orthogonal trajectory through the same point (x, y) then makes an angle φ + π/2 with the x-axis. Since

1tan(φ + π/2) = − cot φ = −

tan φ

dy dx dxand since the slope of the curve is

dx = tan φ, we should replace

dy by −

dy in the differential

equation for the original family to get the differential equation for the orthogonal trajectories.

Example 4.3. We consider the family of circles

(4.6) x 2 + y 2 = cx

tangent to the y axis. By differentiating (4.6) and by eliminating c, we obtain a differential equation

dy 2 dyx 2 + y 2 = 2x + 2xy

dx, or y − x 2 = 2xy

dx that the family of curves (4.6) satisfies. Replace dy/dx by −dx/dy we get the equation of the orthogonal trajectories

dx y 2 − x 2 = −2xy

dy.

Lecture 4 2 18.034 Spring 2009

We write it in differential form as

2xydx − x 2dy + y 2dy = 0.

Multiplying by 1/y2 then gives∗ � �2x

d + dy = 0, y

2xand hence, + y = c. We arrange it into

y

x 2 + y 2 = cy,

which represents a family of circles tangent to the x-axis. Although the analytical steps require x = 0� , y = 0� , and y2 =� x2, the final result is valid without

these restrictions. The quantity c defined by x2 + y2 = cx is constant on the original curves of the family, but not

on the orthogonal trajectories. That is why c must be eliminated in the first step.

Exercise. Show that the orthogonal trajectories of the family of geometrically similar, coaxial el­lipses

x 2 + my 2 = c, m > 0 mare given by y = ±|x| .

Exercise. Show that the solution curves of any separable equation y� = f(x)g(y) have as orthogo­nal trajectories the solution curves of the separable equation y� = −1/f(x)g(y).

∗This procedure makes the equation exact and the solution is defined implicitly. The factor 1/y2 is called a integrating factor. We will study exact differential equation more systematically later.

Lecture 4 3 18.034 Spring 2009